June 2022 Paper 3 Q7
7 In this question you must show detailed reasoning.
Given also that \(m\) is a negative integer, find this value of \(\theta\), correct to 3 significant figures. [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(m\sec\theta + 3\cos\theta = 4\sin\theta\) \(\left(\Rightarrow m\sec\theta + \dfrac{3}{\sec\theta} = 4\sin\theta\right)\) \(m\sec^2\theta + 3 = 4\sin\theta\sec\theta\) | M1 | 2.1 |
| \(m(1 + \tan^2\theta) + 3 = 4\tan\theta\) | M1 | 1.1 |
| \(m + m\tan^2\theta + 3 = 4\tan\theta\) \(\Rightarrow m\tan^2\theta - 4\tan\theta + (m + 3) = 0\) | A1 | 2.2a |
| [3] |
Notes
M1: Or for \(m\sec\theta + 3\cos\theta = 4\sin\theta\) \(\left(\Rightarrow \dfrac{m\sec\theta}{\cos\theta} + 3 = 4\dfrac{\sin\theta}{\cos\theta}\right)\) \(m\sec^2\theta + 3 = 4\tan\theta\)
The first M mark is for a valid method arriving at a three term equation containing \(\sec^2\theta\)
Squaring each individual term of the original equation scores no marks
M1: Correctly uses the identity \(1 + \tan^2\theta \equiv \sec^2\theta\) to obtain an equation in \(\tan\theta\) only
A1: AG so sufficient working must be shown
A0 if angle missing from any trigonometric terms
| Scheme | Marks | AO |
|---|---|---|
| DR \(\Delta = (-4)^2 - 4m(m + 3)\) | M1* | 3.1a |
| As the quadratic equation in tan has only one solution for \(\theta\) in the given interval (and as the range of tan in the given interval is all non-zero real values) this implies that the given equation must only have one real root and therefore \((-4)^2 - 4m(m + 3) = 0 \Rightarrow m^2 + 3m - 4 = 0\) | M1dep* | 1.1 |
| \((m + 4)(m - 1) = 0 \Rightarrow m = -4\) only as \(m\) is a negative integer | A1 | 1.1 |
| \(m = -4 \Rightarrow (2\tan\theta + 1)^2 = 0\) so \(\tan\theta = -0.5\) | M1 | 1.1 |
| \(\theta = 2.68\) (3 sf) | A1 | 2.4 |
| [5] |
Notes
M1*: Considers discriminant of given quadratic equation in tan (\(c\) must be two terms) to get an expression in \(m\) only. M0 for embedded discriminant in quadratic formula unless explicitly considered
Allow \(4^2 - 4m(m + 3)\)
M1dep*: Sets their discriminant equal to zero and obtains an expanded three-term quadratic in \(m\)
Reasoning for setting the discriminant equal to zero is not required for this mark
A1: State or imply \(m = -4\) only
M1: Uses their negative integer value of \(m\), and solves the equivalent of their three term quadratic equation in tan, to obtain (at least) \(\tan\theta = k\), where \(k\) is non-zero - dependent on both previous M marks
Allow \(-4\tan^2\theta - 4\tan\theta - 1 = 0 \Rightarrow \tan\theta = -0.5\) for this mark
If no method shown for solving their quadratic, then award this mark if the solution is correct for their quadratic
A1: For full marks must explain why the discriminant should be set equal to zero – must say that as there is only one value of \(\theta\) or \(\tan\theta \Rightarrow \Delta = 0\) (as a minimum must see explicit mention of ‘one’ together with ‘\(\theta\)’ or ‘\(\tan\theta\)’ for this mark). Allow awrt 2.68
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